Question

If \(2f(x)-3f\!\left(\dfrac{1}{x}\right)=x^2,\; x\neq 0\), then \(f(2)=\)

• A. \(-\dfrac{7}{4}\)
• B. \(\dfrac{5}{2}\)
• C. \(-1\)
• D. \(2\)

Hint:

For functional equations involving \(f(x)\) and \(f\!\left(\frac{1}{x}\right)\):
Replace \(x\) by \(\dfrac{1}{x}\) to form a system of equations
Solve the simultaneous equations algebraically

Answer 1

The Correct Option is A

Step 1: Given, \[ 2f(x)-3f\!\left(\frac{1}{x}\right)=x^2 \qquad (1) \] Replace \(x\) by \(\dfrac{1}{x}\): \[ 2f\!\left(\frac{1}{x}\right)-3f(x)=\frac{1}{x^2} \qquad (2) \]
Step 2: Write equations (1) and (2) in standard form: \[ 2f(x)-3f\!\left(\frac{1}{x}\right)=x^2 \] \[ -3f(x)+2f\!\left(\frac{1}{x}\right)=\frac{1}{x^2} \]
Step 3: Multiply equation (1) by \(2\) and equation (2) by \(3\): \[ 4f(x)-6f\!\left(\frac{1}{x}\right)=2x^2 \] \[ -9f(x)+6f\!\left(\frac{1}{x}\right)=\frac{3}{x^2} \]
Step 4: Add the above equations: \[ -5f(x)=2x^2+\frac{3}{x^2} \] \[ f(x)=-\frac{2x^2+\frac{3}{x^2}}{5} \]
Step 5: Substitute \(x=2\): \[ f(2)=-\frac{2(2)^2+\frac{3}{(2)^2}}{5} =-\frac{8+\frac{3}{4}}{5} =-\frac{\frac{35}{4}}{5} =-\frac{7}{4} \]